Extracting non-contiguous array indices from Scala Array into new Array

Question

Consider the following Scala Array x definition

scala> val x = Array(10, 32, 45, 54, 44, 37)
x: Array[Int] = Array(10, 32, 45, 54, 44, 37)

and an index selector, which is provided, and is guaranteed not to include out of bounds index values:

scala> val selector = Seq(0, 3, 5)
selector: Seq[Int] = List(0, 3, 5)

In Scala, what is a concise way to use selector in combination with x to select only those indices of x which match the desired indices in selector to produce a new result Array? Specifically, result is expected as:

result: Array(10, 54, 37)

Any help is appreciated. Thank you.

Dima · Accepted Answer · 2017-10-24 16:32:01Z

5

Event shorter :)

 import scala.collection.breakOut
 val result: Array[Int]= selector.map(x)(breakOut)

.map takes a parameter of type Function1[Int,_], and when we pass x there, the compiler is smart enough to figure out, that what me mean is x.apply, which is a method Int => Int, and it uses eta-expansion to convert it into a function, so that we don't have to spell it out explicitly, creating an anonymous function like .map { idx => x(idx) }.

breakOut is a collection factory, that tells .map to return the result as Array rather than List (because the expected return type on the LHS is Array). This avoids building an intermediate collection and then copying it into an array in (.map(x).toArray)

edited Oct 24, 2017 at 16:32

answered Oct 24, 2017 at 16:24

Dima

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1 Comment

Jesús Zazueta Over a year ago

I had no idea about breakOut. Short, simple, and I learned something new today. Thank you for the insight.

Brian · Accepted Answer · 2017-10-24 16:18:54Z

1

Access x with each selector using map.

scala> selectors.map(s => x(s)).toArray
res1: Array[Int] = Array(10, 54, 37)

answered Oct 24, 2017 at 16:18

Brian

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